Two particles of masses 1 kg and 3 kg have position vectors `2hati+3hatj+4hatk` and `-2 hati +3 hatj-4hatk` respectively. The centre of mass has a position vector

Options

A. `hati + 3hatj - 2hatk`
B. `-hati - 3 hatj - 2hatk`
C. `-hati + 3hatj + 2hatk`
D. `-hati + 3hatj - 2hatk`

Answer

D

Explanation

Here `m_(1) = 1` kg , `m_(2) =` 3 kg
`vecr_(1) = 2hati + 3hatj + 4hatk , vecr_(2) = -2hati + 3hatj - 4hatk`
The position vector of centre of mass is
`vecR_(CM) = (m_(1) vecr_(1) + m_(2) vecr_(2))/(m_(1) + m_(2)) = ((1) (2 hati + 3hatj + 4hatk) + (3) (-2hati + 3hatj - 4hatk))/(1 + 3)`
`= (2 hati + 3hatj + 4hatk - 6hati + 9hatj - 12 hatk)/(4) = (-4hati + 12hatj - 8hatk)/(4) = - hati + 3hatj - 2hatk`